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Transcript
```An Introduction to Stress and
Strain
Lecture 9/14/2009
GE694 Earth Systems Seminar
Some definitions:
• Body force == A force that acts throughout
an object. Example: gravity is a body force
because gravity affects every every single
atom in an object in the same way.
• Pressure and stress == force per unit area
on the surface of an object. Example: air
pressure acts on the surface of one’s body.
Pressure and stress can be defined on
imaginary surfaces as well as on real
surfaces.
Some mathematics:
^
• A force F acts in some direction n with
magnitude |F|. It can be described by a
vector F=(Fx, Fy, Fz).
y
Fa
Fb
Fa is a stronger force than Fb,
and it acts in a different
direction.
x
F
Fy
Fx
• A stress nm is a force that acts in some
direction n^ with magnitude | nm | on a
surface where a normal vector to the
^
surface points in direction m.
The
complete description of stress at a point
is described by a matrix of 9 values.
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Stress
Intensity of Force
Stress = Force /Area
• Not all nine components of a stress tensor
(matrix) are independent. Since xy=yx,
yz=zy, and xz=zx, there are only 6
independent values.
• A stress tensor can be divided into two
tensors. One tensor is the isotropic tensor,
which is also called the hydrostatic stress
tensor. The other tensor is the deviatoric
stress tensor.
Isotropic
stress (pressure)
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(



p



p
)
Deviatoric stress
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(
'xx 'xy 'xz
’yx ’yy ’yz
’zx ’zy ’zz
)
• Normal stresses -- xx, yy and zz. The
force is oriented in the same direction as the
normal to the surface. Normal stresses can
compress or stretch an object, changing its
volume. Angles cannot be changed by
normal stresses. (Figure A)
• Shear stresses -- xy, xz, yz. The force is
oriented parallel to the surface. Shear
stresses can deform objects by changing
angles in the object (like shearing a deck of
cards). Volume cannot be changed by shear
stresses. (Figure B)
• It is very difficult to determine the absolute
value of stress in the Earth. However, there
are many ways that the relative values of
stress in different directions can be estimated.
It is also possible to estimate the changes of
stress with time.
• Stresses can cause objects to deform through
elastic behavior (spring-like behavior) or
through fluid flow. High values of shear
stresses can cause brittle objects to fracture.
• Strain nm is a deformation of an object
due to an applied stress. The complete
description of strain at a point is
described by a matrix of 9 values.
Strain refers to a change in length per
unit length (L/L ) or to a change in
angles due to an applied stress.
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Shear strains (describe
changes in angles)
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Normal strains (describe length and
volume changes)
Normal strains (describe length and
volume changes)
Dilatation=change in
volume per unit volume
Shear strains (describe
changes in angles)
• Strains can be found by taking spatial derivatives of
displacements. If a displacement changes with
position, then there is a strain. Let wx, wy and wz be
the displacements in the x, y and z directions,
respectively. Then the following expressions show
how elements of the strain tensor are computed from
spatial derivatives of displacements.
Shear strains (describe
changes in angles)
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Normal strains (describe length and
volume changes)
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Dilatation=change in volume per unit volume
• For infinitesimal strains, not all nine components of a
strain tensor (matrix) are independent. Since xy=yx,
yz=zy, and xz=zx, there are only 6 independent
values.
• A strain tensor can be divided into two tensors. One
tensor is the isotropic tensor, which describes volume
changes. The other tensor is the deviatoric strain
tensor, which describes angle changes (shear).
Isotropic strain (volume
change)QuickTime™ and a
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(
e



e



e
)
(e is the dilatation divided by 3)
Deviatoric strain (shear)
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(
'xx 'xy 'xz
’yx ’yy ’yz
’zx ’zy ’zz
)
• Since strain is determined by spatial changes
in displacements, it must always be
referenced to the condition when the
displacements were 0. For many deformation
situations, one determines the strain rate, or
the change of strain with time.
Strain rate example
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• The relationship between stress and
strain is described by a “constitutive
law”. There are many constitutive laws,
each one corresponding to a different
kind of material behavior. Some kinds
of material behaviors that have their
own constitutive laws:
– Elastic material
– Newtonian viscous fluid
– Non-Newtonian viscous fluid
– Elastoplastic material
An Introduction to Fluid
Mechanics
Lecture 9/14/2009
GE694 Earth Systems Seminar
Fluid Mechanics
• Fluids flow due to changes in pressure
with distance. Fluids always flow from
areas of high pressure to areas of low
pressure. The amount of fluid flow is
controlled by the pressure gradient over
distance l, defined as (p0-p1)/l, where p0
and p1 are the pressure values
separated by distance l. The flow of a
fluid depends on the pressure gradient,
the properties of the fluid, drag
conditions that affect the flow, and the
existence or absence of turbulence.
A Simple Fluid Flow Problem:
Channel Flow
• The flow of a fluid in a channel illustrates
some basic properties of fluid flows. In this
problem, there is a flow through a channel
where one boundary of the flow is
stationary and the other moves with the
flow.
This could
represent the
flow of water in a
river.
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Some basic fluid dynamics
concepts:
• “Newtonian” fluid flow law:
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Note: Equation 6-1 is a
constitutive law for a
Newtonian viscous fluid.
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u is the velocity of the fluid
(displacement with time), so
du/dy is a shear strain rate.
 is a shear stress acting on
the fluid.  is called the
dynamic viscosity. For a
viscous fluid, shear stress
causes the fluid to flow.
 is the density of the fluid
• The force balance equation (i.e., Newton’s
second law) is also called the equation of
motion. For channel fluid flow being driven
by a pressure gradient (p0-p1)/l, the
equation of motion is:
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units of distance (g is
the acceleration due
to gravity).
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• The solution for the velocity in the channel
flow problem is:
Equation (6-10) is
the equation of
motion for this
problem.
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Here u0 is the
velocity at the freely
moving surface at
the top of the fluid.
• A drawing of the solution to this problem is:
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Note how the flow must go to 0 at the
base, since a viscous fluid in contact with a
solid boundary must have the same
velocity as the boundary. The solid
boundary base acts to resist the flow, and
the viscous flow imposes a stress on the
top of the solid base.
• Simple variations on this problem:
The top plot corresponds to a
situation where there is no
of the flow is being driven at
velocity u0.
The bottom plot corresponds to
a situation where there is a
of the flow as well as the
bottom of the flow have no
velocity.
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Pipe Flow Problem:
• This problem is similar to the channel flow
problem.
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• Laminar versus turbulent flow:
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Note: The previous solutions that we have looked at all
assume laminar flows.
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• The general properties of fluid flows can be estimated
using dimensionless variables:
Theory of Fluid Flow
• The quantity that is normally measured in
fluid flow problems is the fluid velocity. Let
velocity be a vector (u, v, w) where u is the
velocity in the x direction, v is the velocity
in the y direction, and w is the velocity in
the z direction. For a steady flow, the
amount of fluid flowing into any volume is
the same as the amount of fluid flowing out
of any volume (conservation of mass).
This is described mathematically by the
“continuity equation”.
Continuity Equation:
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Note: The above is a 2-dimensional form of the continuity
equation for an imcompressible fluid. In 3-D, the continuity
equation is
(u/x)+(v/y)+(w/z)=0
Equation of Continuity
Compressible or
Incompressible Fluid Flow
Most liquids are nearly incompressible; that is, the
density of a liquid remains almost constant as the
pressure changes.
To a good approximation, then, liquids flow in an
incompressible manner.
In contrast, gases are highly compressible. However,
there are situations in which the density of a flowing gas
remains constant enough that the flow can be considered
incompressible.
Setting up and solving general
fluid flow problems: A simple
example
• The following is a simple 2-D example of
how a fluid flow problem is set up to find
the differential equations that need to be
solved. In this simple example, there is a
steady-state flow (it does not change with
time) in the x and y directions. Gravity acts
in the y direction.
• The starting equations:
The constitutive law for a Newtonian fluid (in 2-D), including
the pressure that drives the flow
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(xy=xy)
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The continuity equation for an incompressible fluid in 2-D is
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Combine the constitutive law and the continuity law
equations to get the force-balance equations (i.e., Newton’s
second law) for this simple flow. The result is a simple form
of the “Navier-Stokes Equation”:
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• Solving the equations:
Equation pairs 6-64 and 6-65 or 6-66 and 6-67 are the ones
we want to solve for the flow (i.e., the velocities u and v). A
common way to solve these is to solve for a stream function
where
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The solutions for the stream function, and hence for u and v,
can only be completely determined once the boundary and
initial conditions are specified. After the solution for the
stream function is found, the velocities u and v are found by
taking the above derivatives of the stream function.
Streamline Flow
When the flow is steady, streamlines are often used to
represent the trajectories of the fluid particles. Streamlines
are defined by the stream function .
A streamline is a line drawn in the fluid such that a tangent
to the streamline at any point is parallel to the fluid velocity
at that point.
Steady flow is often called streamline flow.
(a) In the steady flow of a liquid, a colored dye reveals the
streamlines. (b) A smoke streamer reveals a streamline
pattern for the air flowing around this pursuit cyclist, as he
tests his bike for wind resistance in a wind tunnel.
Stokes Flow
• The problem of a solid spherical object being
driven up or down in a fluid due to buoyancy
forces (gravity and density differences) is
known as Stokes flow. The solution can be
applied to many different kinds of problems in
the geosciences.
• This problem is best solved in spherical
coordinates, i.e. (, .
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• Here is the geometry of the problem
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The equations of motion are
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The continuity equation for an incompressible fluid is
• The starting equations:
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The continuity equation and the equations of motion can be
manipulated into the forms
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We look for solutions of the form
After solving the equations and applying the boundary
conditions, the final solutions for the velocity components are
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There is drag on the sphere, which is a combination of the
pressure drag Dp and the viscous drag Dv. These two drags are
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```
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